# Limit of trigonometric function involving floor function

I am trying to find the limit of the sequence $$\frac{\sin(2\pi k\lfloor (n-1)/2\rfloor/n)}{2\sin(\pi k/n)}$$ as $n\to\infty$, where $k\ne0$ is a fixed integer and $\lfloor\cdot\rfloor$ is the floor function.

The numerator and the denominator go to $0$ as $n\to\infty$ (we have that $q_n/n\to1/2$ as $n\to\infty$). Since $\sin(x)/x\to1$ as $x\to0$, we see that $\sin(\pi k/n)\sim\pi k/n$ as $n\to\infty$. However, I do not see a way to use the same fact to establish the rate of convergence of the numerator.

It seems that the limit is either $1$ or $-1$ depending on the parity of $k$.

Thank you very much!

• I don't think the limit exists. For the subsequence of odd numbers $n$, the limit evaluates to $\dfrac{(-1)^{k+1}}2$ whereas for the subsequence of even numbers $n$, the limit evaluates to $(-1)^{k+1}$ which aren't equal. – Prasun Biswas Sep 12 '17 at 13:05
• @PrasunBiswas I see. Indeed, the limit does not exist. Thanks for pointing that out! – Cm7F7Bb Sep 12 '17 at 13:25